Problem: Chubby Yang is studying linear equations right now. He came up with a nice problem. In the problem you are given an n × n matrix W, consisting of integers, and you should find two n × n matrices A and B, all the following conditions must hold: 

  * Aij = Aji, for all i, j (1 ≤ i, j ≤ n); 
  * Bij = - Bji, for all i, j (1 ≤ i, j ≤ n); 
  * Wij = Aij + Bij, for all i, j (1 ≤ i, j ≤ n). 



Can you solve the problem?

Input

The first line contains an integer n (1 ≤ n ≤ 170). Each of the following n lines contains n integers. The j-th integer in the i-th line is Wij (0 ≤ |Wij| < 1717).

Output

The first n lines must contain matrix A. The next n lines must contain matrix B. Print the matrices in the format equal to format of matrix W in input. It is guaranteed that the answer exists. If there are multiple answers, you are allowed to print any of them.

The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 4.

Examples

Input

2
1 4
3 2


Output

1.00000000 3.50000000
3.50000000 2.00000000
0.00000000 0.50000000
-0.50000000 0.00000000


Input

3
1 2 3
4 5 6
7 8 9


Output

1.00000000 3.00000000 5.00000000
3.00000000 5.00000000 7.00000000
5.00000000 7.00000000 9.00000000
0.00000000 -1.00000000 -2.00000000
1.00000000 0.00000000 -1.00000000
2.00000000 1.00000000 0.00000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
const int maxn = 170 + 10;
int n;
int w[maxn][maxn];
void solve() {
  cin >> n;
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (n); j++) cin >> w[i][j];
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (n); j++) {
      cout << fixed << setprecision(9) << (w[i][j] + w[j][i]) / 2.0
           << " \n"[j == n - 1];
    }
  for (int i = (0); i < (n); i++)
    for (int j = (0); j < (n); j++) {
      cout << fixed << setprecision(9) << (w[i][j] - w[j][i]) / 2.0
           << " \n"[j == n - 1];
    }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}